Trigonometric function with a theoretical scenario used to find missing variables?

Question

For my pre-calculus class I am given a theoretical scenario and I am tasked with finding the different time(s) an object within the scenario will be "x" inches above equilibrium.

The prompt given to me is:

Suppose that a block is attached to a spring and stretched 6 inches below its equilibrium position, then released. In 2 seconds, the block reaches a maximum height of 6 inches above equilibrium and then returns to its original position 6 inches below equilibrium. Assuming the block continues to oscillate in this manner, the function d(t) = −6 cos(πt) gives the displacement of the block t seconds after it is released. At what times during the first two seconds will the block be 4 inches above equilibrium? (Round your answers to the nearest hundredth of a second.)

What I know for certain is:

• The variable t represents time
• When t = 2 the block reaches the maximum height 6 inches above equilibrium
• After it reaches equilibrium, it returns to its original position 6 inches below equilibrium
• The function d(t) = -6cos(πt) gives the displacement of the block after t seconds after released.

What I have to find is: At what times during the first two seconds will the block be 4 inches above equilibrium? I am unsure how I should go upon finding the requested times.

The only way that I can think of is simply trial-and-error by entering various times into the formula. While this is one way to do it technically, I feel this is not the proper way to do it, so what is?

I did however enter t = 1 into the formula and I got 6, which seems correct, as the prompt states "In 2 seconds, the block reaches a maximum height of 6 inches above equilibrium and then returns to its original position 6 inches below equilibrium."

By solving the equation as 4 = -6 cos(pit) I got the answer 0.7323 which does solve to 4. I feel like I am doing something wrong or approaching this question incorrectly. Do you mind elaborating on what I am doing wrong and what I should do and the steps necessary?

Answer 1

You are taking the correct approach and solving $$-6\cos(\pi t)=4\Rightarrow\cos(\pi t)=-\frac 23$$

Then, $$\pi t=\pm \arccos\left(-\frac 23\right)+n\cdot2\pi$$ where $n$ is an integer.

Choose $n$ to obtain the first two positive times $t$.

Therefore you have $$t=\frac {1}{\pi}\arccos\left(-\frac 23\right)$$ And $$t=-\frac {1}{\pi}\arccos\left(-\frac 23\right)+2$$